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Question-163481

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Question Number 163481 by smallEinstein last updated on 07/Jan/22
Answered by Mathspace last updated on 07/Jan/22
Ψ_n =∫_0 ^∞ cos(x^n )logx dx letf(a)=∫_0 ^∞ x^a cos(x^n )dx f^′ (a)=∫_0 ^∞ x^a logx cos(x^n )dx ⇒f^′ (0)=∫_0 ^∞ cos(x^n )logx dx f(a)=Re(∫_0 ^∞ x^a e^(ix^n ) dx) ∫_0 ^∞ x^a e^(ix^n ) dx let ix^n =−t⇒x^n =it ⇒ x=(it)^(1/n) =e^((iπ)/(2n)) t^(1/(n )) ⇒ ∫_0 ^∞ x^a e^(ix^n ) dx=e^((iπ)/(2n)) ∫_0 ^∞ e^((iπa)/(2n)) t^(a/n) (1/n)t^((1/n)−1) e^(−t) dt =(1/n)e^(((iπ)/(2n))(a+1)) ∫_0 ^∞ t^(((a+1)/n)−1) e^(−t) dt =(1/n)Γ(((a+1)/n))e^(((iπ)/(2n))(a+1)) ⇒f(a)=(1/n)Γ(((a+1)/n))cos(((π(a+1))/(2n))) ⇒f^′ (a)=(1/n^2 )Γ^′ (((a+1)/n))cos(((π(a+1))/(2n))) −(π/(2n^2 ))sin(((π(a+1))/(2n)))Γ(((a+1)/n)) Ψ_n =f^′ (0)=(1/n^2 )Γ^′ ((1/n))cos((π/(2n))) −(π/(2n^2 ))sin((π/(2n)))Γ((1/n)) Ψ_4 =(1/(16))Γ^′ ((1/4))cos((π/8))−(π/(32))Γ((1/4))sin((π/8)) =∫_0 ^∞ cos(x^4 )logx dx
Ψn=∫0∞cos(xn)logxdxletf(a)=∫0∞xacos(xn)dxf′(a)=∫0∞xalogxcos(xn)dx⇒f′(0)=∫0∞cos(xn)logxdxf(a)=Re(∫0∞xaeixndx)∫0∞xaeixndxletixn=−t⇒xn=it⇒x=(it)1n=eiπ2nt1n⇒∫0∞xaeixndx=eiπ2n∫0∞eiπa2ntan1nt1n−1e−tdt=1neiπ2n(a+1)∫0∞ta+1n−1e−tdt=1nΓ(a+1n)eiπ2n(a+1)⇒f(a)=1nΓ(a+1n)cos(π(a+1)2n)⇒f′(a)=1n2Γ′(a+1n)cos(π(a+1)2n)−π2n2sin(π(a+1)2n)Γ(a+1n)Ψn=f′(0)=1n2Γ′(1n)cos(π2n)−π2n2sin(π2n)Γ(1n)Ψ4=116Γ′(14)cos(π8)−π32Γ(14)sin(π8)=∫0∞cos(x4)logxdx

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